TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13

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The dual family of a finite basis is a basis of the dual space, with the same dimension

Statement

If $B = (b_{1}, \dots, b_{n})$ is a basis of a finite-dimensional $F$ -vector space $V$ , then its dual family $B^{*} = (b_{1}^{*}, \dots, b_{n}^{*})$ is a basis of $V^{*}$ . Consequently $\dim V^{*} = \dim V = n$ .

Facts & Assumptions

Given: A finite basis $B = (b_{1}, \dots, b_{n})$ of $V$ and its dual family.

[L1]

The dual functionals satisfy $b_{i}^{*} (b_{j}) = δ_{i j}$ (The dual family $(b^{*})_{b \in B}$ associated to a Hamel basis $B$ , defined by $b^{*} (c) = δ_{b c}$ ).

[L2]

Every $v \in V$ has unique coordinates $v = \sum_{j = 1}^{n} a_{j} b_{j}$ (A finite list $v : n \to V$ is an ordered basis if and only if every $x \in V$ equals $\sum_{i < n} λ_{i} v_{i}$ for exactly one $λ : n \to F$ ; those scalars are the coordinates of $x$ in that ordered basis).

[L3]

The dimension of a finite-dimensional space is the cardinality of any finite basis (Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis).

Proof

technique · direct

1.1L1algebra

If $\sum_{i = 1}^{n} c_{i} b_{i}^{*} = 0$ , evaluation at $b_{j}$ and [L1] give $c_{j} = 0$ for every $j$ ; hence the dual family is linearly independent.

1.2L1L2algebra

For $f \in V^{*}$ , set $g = \sum_{i = 1}^{n} f (b_{i}) b_{i}^{*}$ . If $v = \sum_{j} a_{j} b_{j}$ as in [L2], then $g (v) = \sum_{j} a_{j} f (b_{j}) = f (v)$ , so $f = g$ and the dual family spans $V^{*}$ .

2.1step 1.1step 1.2L3∎

Steps 1.1 and 1.2 make $B^{*}$ a basis, and [L3] gives $\dim V^{*} = n = \dim V$ . For $n = 0$ , both sums are empty and the unique functional on the zero space is zero, so the same proof applies.

Depends on

Dependency tree · 2 levels

Sources

H. Pinkham, Linear Algebra, Chapter 6